Exercise 3.1.4 (Sets are partially ordered by set inclusion)

Proof.

1.

Let $x$ be an arbitrary element of $A$. Since every object that is contained in $A$ is also contained in $B$, we in particular have $x\in B$. Since every object that is contained in $B$ is also contained in $C$, we in particular have $x\in C$. Since our choice of $x\in A$ was arbitrary, we have managed to demonstrate that $$\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A⟹x\in C,$$

as desired.

2.

Since $A$ is a subset of $B$, every element of $A$ is contained in $B$, i.e, $$\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A⟹x\in B.$$

Since $B$ is a subset of $A$ as well, every element of $B$ is contained in $A$, i.e,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in B⟹x\in A.$$

In other words, every element of $A$ is an element of $B$ and vice versa:

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in B⟺x\in A.$$

By Definition 3.1.4, the above means that $A=B,$ as desired.

3.

Remark 1. What does it mean for $A$ to be a proper subset of $B?$ We use the laws of logic to expand on the following definition:

$$A\subseteq B\text{AND}A\ne B.$$

The first condition means that every element of $A$ is an element of $B$ while the second condition means that there exists an element of one set that is not contained in the other one. More formally, by the definition of subsets and the remark in the Exercise 3.1.2, we have:

$$\left[\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A⟹x\in B\right]\text{AND}\left[(\exists <!--\; FUNCTION\; APPLICATION\; -->y:y\in B\text{AND}y\notin A)\text{OR}(\exists <!--\; FUNCTION\; APPLICATION\; -->y:y\in A\text{AND}y\notin B)\right].$$

Notice that the fact that every element of $A$ is contained in $B$ is incompatible with the possibility that there is an element of $A$ that is not contained in $B$. In other words, $(y\in A\text{ AND }y\notin B)$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A⟹x\in B$ cannot be true at once. Therefore, the definition reduces to:

$$\left[\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A⟹x\in B\right]\text{AND}\left[\exists <!--\; FUNCTION\; APPLICATION\; -->y:y\in B\text{AND}y\notin A\right].$$

Thus, every element of $A$ is an element of $B$ but there is an element of $B$ which is not an element of $A$.

By the remark above, we must demonstrate that $A$ is a subset of $C$ yet there exists an element of $C$ which is not an element of $A$.
We first prove $A\subseteq C$. By definition, we have $A\subseteq B$ and $B\subseteq C$. By the first part of this exercise, we thus get $A\subseteq C$.
We now prove that $A\ne C.$ By the remark above, we have an element $y$ of $B$ which is not an element of $A$:

$$\exists <!--\; FUNCTION\; APPLICATION\; -->y:y\in B\text{AND}y\notin A.$$

By the remark again, every element of $B$ is an element of $C$:

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in B⟹x\in C.$$

Because all elements of $B$ are element of $C$, in particular, $y\in C.$ Thus, we can conclude that there exists an element of $C$ which is not contained in $A,$ and therefore, $A\ne C,$ as desired.